Help pleaseeeeeeeeeeeeee

Gala2k [10]

First you have to get the same variables on one side so you’d subtract 5.1w from both sides, making the equation -0.6w = -30

then you want to single out the variable, diving -0.6 on both sides making the new equation and the answer w = 50

hope this helps!

5 0

4 years ago

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The value of n needed to make the equation true.<br> n+19=40

Schach [20]

Answer:

n = 21

Step-by-step explanation:

Given

n + 19 = 40 ( subtract 19 from both sides )

n = 40 - 19 = 21

6 0

4 years ago

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How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago

Step by step how to get 22.04 times 2.8 with work

Bumek [7]

Answer:

61.712

Step-by-step explanation:

Multiply .8x.04

Multiply .8x.0

Multiply .8x2

Than add a zero under the answer of .8x.04

Multiply 2x.04

Multiply 2x.0

Multiply 2x2

Multiply 2x 2

4 0

3 years ago

X+4 divided by x^2+16<br><br> Why can't it be simplified?

KonstantinChe [14]

Let's try to simplify x^2 + 16. It's a sum of two squares:

x^2 + 16 = 0

x^2 = -16

The problem is, we can't take a square root of a negative. This is where imaginary numbers come in.

Remember that square roots have a plus or minus symbol outside:

±√-16 = ±4i

Our two roots are 4i and -4i. Therefore, the trinomial simplifies to:

(x + 4i)(x - 4i)

If we attempt to divide x + 4 by these two binomials, we will find that 4 and 4i are not like terms. Therefore, we can't simplify this expression.

8 0

3 years ago